Add algorithm to find Two unique numbers using bit manipulation with docsets

bit_manipulation/fin_two_unique_numbers.py

@@ -0,0 +1,59 @@
+# Reference : https://www.geeksforgeeks.org/dsa/find-two-non-repeating-elements-in-an-array-of-repeating-elements
+def find_two_unique_numbers(arr: list[int]) -> tuple[int, int]:
+    """
+    Given a list of integers where every element appears twice except for two numbers,
+    find the two numbers that appear only once.
+
+    this method returns the tuple of two numbers that appear only once using bitwise XOR
+    and using the property of x & -x to isolate the rightmost bit
+    (different bit between the two unique numbers).
+
+    >>> find_two_unique_numbers([1,2,3,4,1,2])
+    (3, 4)
+
+    >>> find_two_unique_numbers([4,5,6,7,4,5])
+    (6, 7)
+
+    >>> find_two_unique_numbers([10,20,30,10,20,40])
+    (30, 40)
+
+    >>> find_two_unique_numbers([2,3])
+    (2, 3)
+
+    >>> find_two_unique_numbers([])
+    Traceback (most recent call last):
+        ...
+    ValueError: input list must not be empty
+    >>> find_two_unique_numbers([1, 'a', 1])
+    Traceback (most recent call last):
+        ...
+    TypeError: all elements must be integers
+    """
+
+    if not arr:
+        raise ValueError("input list must not be empty")
+    if not all(isinstance(x, int) for x in arr):
+        raise TypeError("all elements must be integers")
+
+    xor_result = 0
+    for number in arr:
+        xor_result ^= number
+
+    righmost_bit = xor_result & -xor_result
+
+    num1 = 0
+    num2 = 0
+
+    for number in arr:
+        if number & righmost_bit:
+            num1 ^= number
+        else:
+            num2 ^= number
+    a, b = sorted((num1, num2))
+    return a, b
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()